## frx 2 years ago 1). $\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }$ 2). $\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ 2x ^{4}+y ^{4} }$ In the first limit I used that $f(x,y)=f(x,0) \rightarrow 0$ and $f(x,y)=f(0,y)\rightarrow0$ But in the second one that becomes totally wrong, since the x-axis goes to zero and y-axis to 1/3. Why should one change from f(o,y)-> 0 to x=y in the second one, how do I recognize when I should set x=y ?

1. experimentX

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2. frx

?

3. experimentX

just a bookmark!!

4. shubhamsrg

Isn't answer to 1st one 1/2 ?

5. frx

Nope it's 0.

6. experimentX

it's a double limit, you have to be a bit careful with it ... since there are multiple ways to come at (0,0), I think there is a theorem .... but currently I am busty posting my own question.

7. shubhamsrg
8. experimentX

says that limit does not exist, try y=x or y=x^2, if you don't get the same limit then limit does not exist.

9. frx

According to the answerkey to Adams Calculus the answer for the first one is 0 and the second one DNE

10. Zarkon

$\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }$ let $$y=mx$$ then $\lim_{x \rightarrow 0} \frac{ x ^{2}(mx) ^{2} }{ x ^{4}+(mx) ^{4} }$ $\lim_{x \rightarrow 0} \frac{ m^2x ^{4} }{ x ^{4}(1+m ^{4}) }$ $\lim_{x \rightarrow 0} \frac{ m^2 }{ 1+m ^{4} }=\frac{ m^2 }{ 1+m ^{4} }$ pick any real number $$m$$